Question

Find '$c$', so that $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$, if $f\left(x\right)=e^x$ at $a=0,b=1$.

Answer 1

$f\left(x\right)=e^x$,
and
$f^{\prime }\left(x\right)=e^x$
put ${}^{\prime }x=c^{\prime }$,
therefore,
$f^{\prime }\left(c\right)=e^c$
from LMVT,
$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
by putting values,
$e^c=\frac{f\left(1\right)-f\left(0\right)}{1-0}$
=> $e^c=\frac{e^1-1}{1}$
taking logarithm with base 'e' on both sides,
$c\ln e=\ln (e-1)$

=> $c=\ln (e-1)$